\(a,\Rightarrow x\left(x+3\right)-\left(x-3\right)\left(x+3\right)=0\\ \Rightarrow\left(x+3\right)\left(x-x+3\right)=0\\ \Rightarrow3\left(x+3\right)=0\Rightarrow x=-3\\ b,A:B=\left(2x^2-x+4x-2\right):\left(2x-1\right)\\ =\left[x\left(2x-1\right)+2\left(2x-1\right)\right]:\left(2x-1\right)\\ =x+2\)

Ta có: \(x+2\sqrt{2}.x^2+2x^3=0\)

\(\Leftrightarrow x\left(1+2\sqrt{2}.x+2x^2\right)=0\)

\(\Leftrightarrow x\left[1^2+2.x\sqrt{2}.1+\left(x\sqrt{2}\right)^2\right]=0\)

\(\Leftrightarrow x\left(1+x\sqrt{2}\right)^2=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\1+x\sqrt{2}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{-1}{\sqrt{2}}\end{cases}}\)

Vậy\(x\in\left\{0;\frac{-1}{\sqrt{2}}\right\}\)

\(x+2\sqrt{2}x^2+2x^3=0\)

\(x\left(1+2\sqrt{2}x+2x^2\right)=0\)

\(x\left(2\sqrt{2}x+1\right)^2=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\2\sqrt{2}x+1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2x\sqrt{2}}\end{cases}}\)

Thôi nhầm chỗ HĐT rồi , bạn làm giống bạn Lê Tài Bảo Châu  nhé ! NHầm

x+2x+......+10x=-165

=>x.(1+2+.......+10)=-165

=>x.55=-165

=>x=-165:55

=>x=-3

x+2x+3x+...+10x = -165

x.(1 + 2 + 3 +...+10)= -165

x.55                       = -165

x                           = -165:55

x                           = -3

x+2\(\times\)x+3\(\times\)x+......+10\(\times\)x=-165

x\(\times\)(1+2+3+....+10)=-165

x\(\times\)55=-165

x=-165\(\div\)55

x=-3

Vậy số cần tìm là -3

Ai tích mình mình tích lại cho.

<=>\(\left(x^3-4x^2\right)+\left(x^2-4x\right)+\left(5x-20\right)=0\)

<=>\(x^2\left(x-4\right)+x\left(x-4\right)+5\left(x-4\right)=0\)

<=>\(\left(x^2+x+5\right)\left(x-4\right)=0\)

Vì \(x^2+x+5>0\)=>x-4=0

<=>x=4

thấy sai sai bạn ạ

\(x^2+2x-10=0\)

\(\Leftrightarrow x^2+2x+1-9=0\)

\(\Leftrightarrow\left(x+1\right)^2-9=0\\\)

\(\Leftrightarrow\left(x+1\right)^2=9\)

\(\Leftrightarrow\left(x+1\right)^2=\pm\sqrt{9}\)

\(\Leftrightarrow\left(x+1\right)^2=\left(\pm3\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3-1\\x=-3-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)

Vậy S={2;-4}

\(x^2\) + 2\(x\) - 10 = 0

\(x^2\) + 2\(x\) + 1 - 11 = 0

(\(x^2\) + 2\(x\) + 1) - 11 = 0

(\(x\) + 1)² - 11 = 0

(\(x\) + 1)² = 11

\(\left[{}\begin{matrix}x+1=\sqrt{11}\\x+1=-\sqrt{11}\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\sqrt{11}-1\\x=-\sqrt{11}-1\end{matrix}\right.\)

Em tách ra thành:

x(1+3+5+...+2021)-x(2+4+...+2020)=2022.

Sau đó giải bình thường.

Chúc em học tốt!

x+2x+3x+...+100x=5500

5500=((1+100)*50)x

=5050

bạn sai đề